In a new system, the unit of mass is made 10 times, the unit of length is made 1/100 times and unit of time is made 10 times. Magnitude of 1J in the new system of unit is `10^x`.What is the value of x ?

Answer

To solve the problem, we need to analyze how the changes in the units of mass, length, and time affect the unit of energy, which is the joule (J). The joule is defined as:

\[ 1 \text{ J} = 1 \text{ kg} \cdot \text{m}^2/\text{s}^2 \]

### Step-by-Step Solution:

1. **Identify the changes in units:**
- The unit of mass (kg) is changed to \(10 \times \text{kg}\).
- The unit of length (m) is changed to \(\frac{1}{100} \text{m}\).
- The unit of time (s) is changed to \(10 \times \text{s}\).

2. **Express the new units:**
- Let the new unit of mass be \(M' = 10 \text{ kg}\).
- Let the new unit of length be \(L' = \frac{1}{100} \text{ m}\).
- Let the new unit of time be \(T' = 10 \text{ s}\).

3. **Substitute the new units into the joule formula:**
\[ 1 \text{ J} = M' \cdot (L')^2 / (T')^2 \]
Substituting the new units:
\[ 1 \text{ J} = (10 \text{ kg}) \cdot \left(\frac{1}{100} \text{ m}\right)^2 / (10 \text{ s})^2 \]

4. **Calculate the new expression:**
- Calculate \((L')^2\):
\[ (L')^2 = \left(\frac{1}{100} \text{ m}\right)^2 = \frac{1}{10000} \text{ m}^2 = 10^{-4} \text{ m}^2 \]
- Calculate \((T')^2\):
\[ (T')^2 = (10 \text{ s})^2 = 100 \text{ s}^2 = 10^2 \text{ s}^2 \]

5. **Substitute these values back into the joule equation:**
\[ 1 \text{ J} = 10 \text{ kg} \cdot 10^{-4} \text{ m}^2 / 10^2 \text{ s}^2 \]

6. **Simplify the expression:**
\[ 1 \text{ J} = 10 \cdot 10^{-4} \cdot 10^{-2} \text{ kg} \cdot \text{m}^2/\text{s}^2 \]
\[ 1 \text{ J} = 10^{1 - 4 - 2} \text{ kg} \cdot \text{m}^2/\text{s}^2 \]
\[ 1 \text{ J} = 10^{-5} \text{ kg} \cdot \text{m}^2/\text{s}^2 \]

7. **Conclusion:**
The magnitude of 1 joule in the new system of units is \(10^{-5}\). Therefore, the value of \(x\) is \(-5\).

### Final Answer:
\[ x = -5 \]